Is it possible to evaluate the inverse of this function, in order to obtain for each $y\in\mathbb R^+$ an explicit value of $f^{-1}(y)$? Thanks in advance!
$f(\delta)=(\frac{1}{Z})\delta^{-\alpha}e^{-\beta d\delta}$
[Function from this article, page 8, item n°4]
Use the inverse function theorem, a function has an inverse in an interval $(a-c,a+c)$, if it is continuously differentiable at $a$.
We take the $\delta$ derivative to obtain $f'(\delta)=-\frac{\alpha}{z}\delta^{\alpha-1}e^{-\beta d\delta}-\frac{\beta d}{z}\delta^{-\alpha}e^{-\beta d\delta}$, and $f'(1)=-\frac{\alpha}{z}e^{-\beta d}-\frac{\beta d}{z}e^{-\beta d}\neq0$, so it has an inverse in a neighbourhood of $1$.
And in fact the derivative is nonzero $\forall y\in\mathbb{R^+}$, so we need not even restrict to a neighbourhood of $1$.
Following up the other post we have:
$$\frac{1}{z}(-\beta d-\alpha f^{-1}(\delta))\delta(f^{-1})'(\delta)=1$$
Let $g(\delta)=f^{-1}(\delta)$, and we have:
$$\int(-\beta d-\alpha g(\delta))dg(\delta)=\int \frac{z}{\delta}d\delta$$
$$\Rightarrow \left(-\beta d g(\delta)-\frac{\alpha}{2\delta}g(\delta)^2\right)=z\ln(\delta)+C\Rightarrow \left(\beta d g(\delta)+\frac{\alpha}{2\delta}g(\delta)^2\right)+z\ln(\delta)+C=0$$
Thus: $$g(\delta)=f^{-1}(\delta)=\frac{-\beta d\pm\sqrt{\beta\delta-2\frac{\alpha}{\delta}(\ln(\delta)+C)}}{\frac{\alpha}{\delta}}$$
Take a look at the Lambert function : http://en.wikipedia.org/wiki/Lambert_W_function The inverse you're looking for is closely related to the Lambert function. However, it does not help much in general computations
If you derive $f$, you'll get:
$f'(\delta)=(\frac{1}{Z})(-\beta d -\alpha\delta ) f(\delta)$
Using it:
$f(f^{-1}(\delta)) = \delta$
$f'(f^{-1}(\delta)) * (f^{-1})'(\delta) = 1$
$(\frac{1}{Z})(-\beta d -\alpha f^{-1}(\delta) ) *\delta * (f^{-1})'(\delta) = 1$
Maybe you can take it from here to solve this equation
